为什么我们使用static_cast为NULL

When I try to study QP/CPP code I came across below line.

当我尝试研究QP/CPP代码时，我遇到了下面一行。

QTimeEvt *t; 
// ...
if (t == static_cast<QTimeEvt *>(0)) {

Why they are doing static_cast of 0? If they want to check NULL we can do that directly right?

为什么他们做的是static_cast (0) ?如果他们想检查NULL我们可以直接做，对吧?

This source code you can find out in

这是你能找到的源代码。

http://www.state-machine.com/qpcpp/qf__time_8cpp_source.html

2 个解决方案

#1

Yeah, that's unnecessary, though it may be mandated by some style guide for "clarity", or it may be there to silence an overzealous static analysis tool.

是的，这是不必要的，尽管它可能是由一些“清晰”的风格指南强制要求的，或者它可能会在那里沉默一个过于热情的静态分析工具。

Of course, nowadays, we'd just write nullptr and leave it at that.

当然，现在我们只要写nullptr就行了。

#2

The idiomatic way to write

习惯写法

QTimeEvt *t; 
// ...
if (t == static_cast<QTimeEvt *>(0)) {

… is

…

QTimeEvt* t; 
// ...
if( !t ) {

Or you can write it as

或者你可以把它写成。

if( not t ) {

… although you can also write

虽然你也可以写作

if( t == nullptr ) {

… or, ¹C++03-style,

……或者,¹c++ 03-style,

if( t == 0 ) {

There's no need for the cast.

不需要演员阵容。

It's just verbiage consuming the reader's time.

这只是在浪费读者的时间。

^{Notes:
¹ If the <stddefs.h> header is included one can write NULL instead of 0 where a nullpointer is required. With a modern implementation NULL might even be defined as nullptr.}

注:如果< stddefs¹。h>标头包含一个可以写入NULL而不是0的空指针。使用现代实现，NULL甚至可以定义为nullptr。

#1